Integral Fail

[Albert Chan]

(08-15-2020 05:20 AM)lrdheat Wrote:  I was integrating from 1 to 3. Do you get the same answer (that would imply symmetry about x=1).

Yes, f(x)=|x-1|^(8/3) have symmetry at x=1. But, for F(x), we need to flip the sign.
Because F(1)=0, integrating from 1 to 3 is same as -1 to 1

F(3) - F(1) = F(1+2) = - F(1-2) = F(1) - F(-1)

FYI, here is an equivalent F(x), using surd:

XCas> F := int(surd(x-1,3)^8,x)
"Temporary replacing surd/NTHROOT by fractional powers"

→ \(\frac{3}{11} \cdot \left(3\mbox{ NTHROOT }(x-1)\right)^{2} \left(x-1\right)^{3}\)             // = \(\frac{3}{11} \cdot \left(3\mbox{ NTHROOT }(x-1)\right)^{11}\)